How do you use a Taylor series to find the derivative of a function?

1 Answer
Sep 27, 2014

One of the benefits of a Taylor series is the ease of differentiation since it can be done term by term. So, the Talyor series

#f(x)=sum_{n=0}^infty{f^{(n)}(a)}/{n!}(x-a)^n#

can be differentiated as

#f'(x)=sum_{n=0}^infty{f^{(n)}(a)}/{n!}[(x-a)^{n}]'#

by Power Rule,

#=sum_{n=1}^infty{f^{(n)}(a)}/{n!}[n(x-a)^{n-1}]#

(Note: #n# starts from #1# since the term is zero when #n=0#)
by simplifying a little further,

#=sum_{n=1}^infty{f^{(n)}(a)}/{(n-1)!}(x-a)^{n-1}#

I hope that this was helpful.